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# proof of argument principle

Since $f$ is meromorphic, $f^{{\prime}}$ is meromorphic, and hence $f^{{\prime}}/f$ is meromorphic. The singularities of $f^{{\prime}}/f$ can only occur at the zeros and the poles of $f$.

I claim that all singularities of $f^{{\prime}}/f$ are simple poles. Furthermore, if $f$ has a zero at some point $p$, then the residue of the pole at $p$ is positive and equals the multiplicity of the zero of $f$ at $p$. If $f$ has a pole at some point $p$, then the residue of the pole at $p$ is negative and equals minus the multiplicity of the pole of $f$ at $p$.

To prove these assertions, write $f(x)=(x\!-\!p)^{n}g(x)$ with $g(p)\neq 0$. Then

${f^{{\prime}}(x)\over f(x)}={n\over x\!-\!p}+{g^{{\prime}}(x)\over g(x)}$ |

Since $g(p)\neq 0$, the only singularity of $f^{{\prime}}\!/\!f$ at $p$ comes from the first summand. Since $n$ is either the order of the zero of $f$ at $p$ if $f$ has a zero at $p$ or minus the order of the pole of $f$ at $p$ if $f$ has a pole at $p$, the assertion is proven.

By the Cauchy residue theorem, the integral

${1\over 2\pi i}\int_{C}{f^{{\prime}}(z)\over f(z)}dz$ |

equals the sum of the residues of $f^{{\prime}}/f$. Combining this fact with the characterization of the poles of $f^{{\prime}}/f$ and their residues given above, one deduces that this integral equals the number of zeros of $f$ minus the number of poles of $f$, counted with multiplicity.

## Mathematics Subject Classification

30E20*no label found*

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